-4
1000
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
Just six numbers... 345, 354, 435, 453, 534 & 543
None.
The first digit of your password can be any of the six numbers. As repetition is allowed so can the second, third and fourth so total possibilities is 6 x 6 x 6 x 6 ie 1296
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
290
if the repetition is allowed the there is 6*6*6 possible ways = 216
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
64 if repetition is allowed.24 if repetition is not allowed.
When the repetition is allowed: 4*4*4 = 64 numbers. Without repetition: 4*3*2 = 24 numbers.
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
There are 504 of them.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
There are 5*5*5 = 125 such numbers.